it's correct!! you get w/ 225 cases decimal. but you calculated still too time to result?
how many digits and many exponents w/ iterations (= n_1^n_2^...^n_k^x) for to evaluate better? do you use maple?

I am 7000 digits and 199 exponents for evaluate, using maple, in only 100 cases decimal.

nuninho1980 Wrote:it's correct!! you get w/ 225 cases decimal. but you calculated still too time to result?
how many digits and many exponents w/ iterations (= n_1^n_2^...^n_k^x) for to evaluate better? do you use maple?

It didn't take that much time at all. The base is very close to 1, which means regular iteration works much better (The natural superlogarithm is very erratic for bases between 1 and eta). Also, I used Mathematica. So I used a 10-term and 15-term regular iteration series (polynomial), compared them, and each took about 3 seconds to evaluate. When they were identical, I assumed all the digits were "significant".

andydude Wrote:It didn't take that much time at all. The base is very close to 1, which means regular iteration works much better (The natural superlogarithm is very erratic for bases between 1 and eta). Also, I used Mathematica. So I used a 10-term and 15-term regular iteration series (polynomial), compared them, and each took about 3 seconds to evaluate. When they were identical, I assumed all the digits were "significant".

Andrew Robbins

slog_b (x) = y
I didn't convert from natural slog to tetration using maple but I tried many times to add new each one of cases decimal of x since x=1 when the result is very close to y=0.5. these tentations last 30-60 minutes.
do you convert from natural slog to tetration?

No, I used regular iteration from the fixed-point p =
1.000000000000000000010000000000000000000100000000000000000001500000000000000000023333333333333333333733333333333333333340408333333333333333463361111111111111113553591269841269841316667063492063492064403294973544973544991509125661375661376019282963213484046824576164483124899791712269571579076787413095423662162995000357635728649265105119539204243709252241024379290515068892945934184599484695347437210742175409191006184949392420206632843923492376014272124055526327911424878272
in the series
where a = 1.00000000000000000001

andydude Wrote:I used regular iteration from the fixed-point p =
1.000000000000000000010000000000000000000100000000000000000001500000000000000000023333333333333333333733333333333333333340408333333333333333463361111111111111113553591269841269841316667063492063492064403294973544973544991509125661375661376019282963213484046824576164483124899791712269571579076787413095423662162995000357635728649265105119539204243709252241024379290515068892945934184599484695347437210742175409191006184949392420206632843923492376014272124055526327911424878272
in the series
where a = 1.00000000000000000001

Andrew Robbins

this method is much faster. thanks!

p+ ... (1-p)^2+ ...
what do you add new last terms, if you want 15 or more terms?